 # More Number Theory Proofs Rosen 1.5, 3.1. Prove or Disprove If m and n are even integers, then mn is divisible by 4. The sum of two odd integers is odd.

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More Number Theory Proofs Rosen 1.5, 3.1

Prove or Disprove If m and n are even integers, then mn is divisible by 4. The sum of two odd integers is odd. The sum of two odd integers is even. If n is a positive integer, then n is even iff 3n 2 +8 is even. n 2 + n + 1 is a prime number whenever n is a positive integer. n 2 + n + 1 is a prime number whenever n is a prime number. |x| + |y|  |x + y| when x,y  R.  3 is irrational.

If m and n are even integers, then mn is divisible by 4. Proof: m and n are even means that there exists integers a and b such that m =2a and n = 2b Therefore mn = 4ab. Since ab is an integer, mn is 4 times an integer so it is divisible by 4.

The sum of two odd integers is odd. This is false. A counter example is 1+3 = 4

The sum of two odd integers is even. Proof: If m and n are odd integers then there exists integers a,b such that m = 2a+1 and n = 2b+1. m + n = 2a+1+2b+1 = 2(a+b+1). Since (a+b+1) is an integer, m+n must be even.

If n is a positive integer, then n is even iff 3n 2 +8 is even. Proof: We must show that n is even  3n 2 +8 is even, and that 3n 2 +8 is even  n is even. First we will show if n is even, then 3n 2 +8 is even. n even means there exists integer a such that n = 2a. Then 3n 2 +8 = 3(2a) 2 + 8 = 12a 2 + 8 = 2(6a 2 + 4) which is even since (6a 2 + 4) is an integer.

If n is a positive integer, then n is even iff 3n 2 +8 is even (cont.). Now we will show if 3n 2 +8 is even, then n is even using the contrapositive (indirect proof). Assume that n is odd, then we will show that 3n 2 +8 is odd. n odd means that there exists integer a such that n = 2a+1. 3n 2 +8 = 3(2a+1) 2 + 8 = 3(4a 2 + 4a + 1) + 8 = 2(6a 2 + 6a + 5) + 1, which is odd. Therefore, by the contrapositive if 3n 2 +8 is even, then n is even.

n 2 + n + 1 is a prime number whenever n is a positive integer. Try some examples: n = 1, 1+1+1 = 3 is prime n = 2, 4+2+1 = 7 is prime n = 3, 9+3+1 = 13 is prime n = 4, 16+4+1 = 21 is not prime and is a counter example. Not true.

n 2 + n + 1 is a prime number whenever n is a prime number. Try some examples: n = 1, 1+1+1 = 3 is prime n = 2, 4+2+1 = 7 is prime n = 3, 9+3+1 = 13 is prime n = 5, 25+5+1 = 31 is prime n = 7, 49+7+1 = 57 is not prime (19*3). Not true.

Prove |x| + |y|  |x + y| when x,y  R. Note: |z| is equal to z if z  0 and equal to -z if z < 0 There are four cases x y  0 <0  0  0 <0 <0

Prove |x| + |y|  |x + y| when x,y  R. Case 1 x,y are both  0 Then |x| + |y| = x + y = |x+y| since both x and y are positive.

Prove |x| + |y|  |x + y| when x,y  R. Case 2 x < 0 and y  0 so |x| + |y| = -x + y If y  -x, then x+y is nonnegative and |x+y| = x+y Since x is negative, -x > x, so that |x| + |y| = -x + y > x+y = |x+y| If y < -x, then |x+y| = -(x+y) = -x + -y. Since y  0, then y  -y, so that |x| + |y| = -x + y  -x + -y = |x+y|

Prove |x| + |y|  |x + y| when x,y  R. Case 3 x  0 and y <0 so |x| + |y| = x + -y If x  -y, then x+y is nonnegative and |x+y| = x+y Since y is negative, -y > y, so that |x| + |y| = x + -y > x+y = |x+y| If x < -y, then |x+y| = -(x+y) = -x + -y. Since x  0, then x  -x, so that |x| + |y| = x + -y  -x + -y = |x+y|

Prove |x| + |y|  |x + y| when x,y  R. Case 4 x,y are both < 0 Then |x| + |y| = -x + - y = -(x+y) = |x+y| Therefore the theorem is true. This is know in mathematics as the Lipschitz condition.

Prove that 3  3 is irrational. Proof (by contradiction): Assume that 3  3 is rational, i.e. that 3  3 = a/b for a,b  Z and b  0. Since any fraction can be reduced until there are no common factors in the numerator and denominator, we can further assume that a and b have no common factors.

Prove that 3  3 is irrational. (cont.) Then 3 = a 3 /b 3 which means that 3b 3 = a 3 so a 3 is divisible by 3. Lemma: When m is a positive integer, then if m 3 is divisible by 3, then m is divisible by 3. (Left as an exercise). By the lemma since a 3 is divisible by 3, then a is divisible by 3. Thus  k  Z  a = 3k.

Prove that 3  3 is irrational. (cont.) Now, we will show that b is divisible by 3. From before, a 3 /b 3 = 3  3b 3 = a 3 = (3k) 3. Dividing by 3 gives b 3 = 9k 3 = 3(3k 3 ). Therefore b 3 is divisible by 3 and from the Lemma, b is divisible by 3.

Prove that 3  3 is irrational. (cont.) But, if a is divisible by 3 and b is divisible by 3, then they have a common factor of 3. This contradicts our assumption that our a/b has been reduced to have no common factors. Therefore 3  3  a/b for some a,b  Z, b  0. Therefore 3  3 is irrational.

Some Other Proof Strategies Rosen 3.1

Backward Reasoning We have used mostly forward reasoning strategies up to now. However, sometimes it is unclear how to proceed from the initial assumptions. Backward reasoning may help-- Motto: If you can’t prove the original proposition, equate it to one you can prove.

Prove  (a,b  Z +, a≠b)[(a+b)/2>  ab] (i.e., the arithmetic mean is always greater than the geometric mean for this universe of discourse.) Backward reasoning proof (a+b)/2>  abOriginal Assumption (a+b) 2 /4> abWhy? (a+b) 2 > 4abWhy? a 2 +2ab+b 2 > 4abWhy? a 2 -2ab+b 2 > 0Why? (a-b) 2 > 0Why?

Prove  (a,b  Z +, a≠b)[(a+b)/2>  ab] Backward reasoning proof (cont.) But,  a,b  Z +, (a-b) 2 > 0 implies a≠b. We can now easily start from a≠b and go backwards to reconstruct the path to prove the original proposition.

Conjecture and Proof A conjecture is a plausible statement that has not been proved. A conjecture may result from recognizing that there are multiple examples for which it is true. For some conjectures, counterexamples are eventually found. However, even if a conjecture is valid for very many examples, this does not usually constitute a valid proof. (Why?)

Conjecture and Proof Sometimes a complex proof is constructed as a series of conjectures that are then proved. Sometimes a proof is found by referring to the proof of a similar problem or class of problem. There are many famous conjectures that are still not proved (or only recently proved).

Fermat’s Last Theorem x n + y n = z n has no solution for x,y,z,n  Z, x,y,z≠0, n>2 This is a conjecture made by Pierre de Fermat (1601-1665), the French mathematician. He wrote in the margin of his copy of the works of Diophantus, an ancient Greek mathematician, that he had a “wondrous proof”, but that it wouldn’t fit in the margin. He then died, leaving no record of the proof!

Fermat’s Last Theorem Attempts at proof over the years led to new fields, such as algebraic number theory. Finally, in 1994, Andrew Wiles provided a correct proof that required hundreds of pages of advanced mathematics.

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